Question

求和：Sn=1+3x+5x^2+.....+(2n-1)X^(n-1)

Answer 1

数列的每一项的公比为x

所以同乘以x，得到

xsn= x+3x^2+5x^3+7x^4+...+(2n-3)*x^(n-1)+(2n-1)x^n ①

因为 sn=1+3x+5x^2+7x^3+9x^4+...+(2n-1)*x^(n-1) ②

②-①得,(1-x)Sn=1+2[x+x^2+x^3+x^4+.....+x^(n-1)]-(2n-1)x^n

(1-x)Sn=1+2[(x-x^n)/(1-x)]-(2n-1)x^n

(1-x)Sn=1+(2x-2x^n)/(1-x)-2nx^n+x^n

(1-x)Sn=1+(2x)/(1-x)-(2x^n)/(1-x)-2nx^n+x^n

(1-x)Sn=1+(2x)/(1-x)+{1-2n-[2/(1-x)}x^n

故：Sn={1+(2x)/(1-x)+{1-2n-[2/(1-x)}x^n}/(1-x)

Answer 2

shab